4) \( x^{2}+\ln (x+y)=1: \) \( \begin{array}{llll}\text { (A) }-2(x+y) & \text { (B) }-2 x(x+y) & \text { (C) }-2 x^{2}-2 y-1 & \text { (D) }-2 x(x+y)-1\end{array} \)

To find the derivative of the equation \( x^{2} + \ln(x+y) = 1 \) with respect to \( x \), we will use implicit differentiation. 1. Differentiate both sides of the equation with respect to \( x \): \[ \frac{d}{dx}(x^2) + \frac{d}{dx}(\ln(x+y)) = \frac{d}{dx}(1) \] 2. The derivative of \( x^2 \) is \( 2x \). 3. For \( \ln(x+y) \), we use the chain rule: \[ \frac{d}{dx}(\ln(x+y)) = \frac{1}{x+y} \cdot \frac{d}{dx}(x+y) = \frac{1}{x+y} \cdot \left(1 + \frac{dy}{dx}\right) \] 4. The derivative of the right side (which is a constant) is \( 0 \). Putting it all together, we have: \[ 2x + \frac{1 + \frac{dy}{dx}}{x+y} = 0 \] 5. Now, we can solve for \( \frac{dy}{dx} \): \[ 2x + \frac{1 + \frac{dy}{dx}}{x+y} = 0 \] Multiply through by \( x+y \) to eliminate the fraction: \[ 2x(x+y) + 1 + \frac{dy}{dx} = 0 \] 6. Rearranging gives: \[ \frac{dy}{dx} = -2x(x+y) - 1 \] Thus, the derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = -2x(x+y) - 1 \] Looking at the options provided, the correct answer is: \[ \text{(D) } -2x(x+y) - 1 \]